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If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become
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Verified Answer
The correct answer is:
half
half
de - Broglie wavelength is given by :
$$
\lambda=\frac{h}{m v}
$$
$$
\begin{aligned}
& \text { K.E. }=\frac{1}{2} m v^2 \\
& v^2=\frac{2 K E}{m} \\
& v=\sqrt{\frac{2 K E}{m}}
\end{aligned}
$$
Substituting this in equation (i)
$$
\lambda=\frac{h}{m} \sqrt{\frac{m}{2 K E}}
$$
$\lambda=h \sqrt{\frac{1}{2 m(\text { K.E. })}}$
i.e. $\lambda \propto \frac{1}{\sqrt{K E}}$
$\therefore$ when $K E$ become 4 times wavelength become $1 / 2$.
$$
\lambda=\frac{h}{m v}
$$
$$
\begin{aligned}
& \text { K.E. }=\frac{1}{2} m v^2 \\
& v^2=\frac{2 K E}{m} \\
& v=\sqrt{\frac{2 K E}{m}}
\end{aligned}
$$
Substituting this in equation (i)
$$
\lambda=\frac{h}{m} \sqrt{\frac{m}{2 K E}}
$$
$\lambda=h \sqrt{\frac{1}{2 m(\text { K.E. })}}$
i.e. $\lambda \propto \frac{1}{\sqrt{K E}}$
$\therefore$ when $K E$ become 4 times wavelength become $1 / 2$.
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